[jaggyd]

Estimate the enthalpy of formation of H2S from S8 (a cyclic molecule) and H2. No mean bond enthalpies are given. please solve it.

[Borek]

Please read the forum rules. You have to show your attempts at solving the question to receive help. This is a forum policy.

[Big-Daddy]

Try drawing the structure of S₈ - it tells you it's cyclical, I may as well tell you it consists purely of single bonds arranged around the 8-member ring in a bent v-shape. So you'll have 8 S-S bonds to break, 8 H-H bonds to break (why 8? try balancing the equation: 8 H₂ + S₈  8 H₂S) and 16 S-H bonds to form.

Now use:

ΔH_r°=Σ(ΔH_BD°[Reactants])-Σ(ΔH_BD°[Products])
ΔH_r°=8·ΔH_BD°[S-S]+8·ΔH_BD°[H-H]-16·ΔH_BD°[S-H]

Luckily it's quite neat to divide through by 8, which is exactly what we need to produce 1 mole of H₂S according to the enthalpy of formation:

ΔH_f°[H₂S]=ΔH_BD°[S-S]+ΔH_BD°[H-H]-2·ΔH_BD°[S-H]

And now it is your turn to make a guess-timate.